\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2014 (2014), No. 220, pp. 1--22.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2014 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2014/220\hfil Quasilinear problems with two parameters] {Quasilinear problems with two parameters including superlinear and gradient terms} \author[M. C. Rezende, C. A. Santos\hfil EJDE-2014/220\hfilneg] {Manuela C. Rezende, Carlos A. Santos} % in alphabetical order \address{Manuela C. Rezende \newline Departamento de Matem\'atica, Universidade de Bras\'ilia, 70910-900 Bras\'ilia, DF - Brasil} \email{manuela@mat.unb.br} \address{Carlos A. Santos \newline Departamento de Matem\'atica, Universidade de Bras\'ilia , 70910-900 Bras\'ilia, DF - Brasil} \email{csantos@unb.br} \thanks{Submitted April 10, 2013. Published October 21, 2014.} \subjclass[2000]{35J92, 35J75, 35B08, 35B09} \keywords{ $p$-Laplacian; entire solutions; superlinear; sublinear; gradient term} \begin{abstract} In this article, we establish conditions for the existence of solutions for a quasilinear elliptic two-parameter problem with dependence on the gradient term in smooth bounded domains or in the whole space $\mathbb{R}^N$. We consider superlinear and asymptotically linear terms. Estimates on the values of two parameters for which the problem have solutions are provided. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \allowdisplaybreaks \section{Introduction} This article concerns the existence of solutions and estimates of the intervals of parameters for which the problem \begin{equation} \label{1} \begin{gathered} -\Delta_p u = a(x)f(u) +\lambda b(x)g(u)+\mu V(x, \nabla u)\quad \text{in } \Omega,\\ u > 0\quad \text{in }\Omega,\quad u=0\quad \text{on } \partial\Omega, \end{gathered} \end{equation} has a solution. Here, $\Delta_p u= \operatorname{div}(| \nabla u |^{p-2} \nabla u)$, $ 10$; $\mu\geq 0$ are real parameters; $f, g:(0, \infty)\to [0,\infty)$; $a, b: \Omega\to[0,\infty )$ with $a, b\neq 0$; $V:\Omega\times\mathbb{R}^N\to [0, \infty)$ are continuous functions satisfying appropriate hypotheses and either $\Omega\subset\mathbb{R}^N$ is a smooth bounded domain or $\Omega=\mathbb{R}^N$. When $\Omega=\mathbb{R}^N$, the condition $u=0$ on $\partial\Omega$ means that $u(x)\to 0$ when $|x|\to\infty$. By a solution of \eqref{1} we mean a function $u=u_{\lambda, \mu} \in C^1(\Omega)\cap C(\overline{\Omega})$, with $u>0$ in $\Omega$, $u=0$ on $\partial\Omega$ and \[ \int_{\Omega} | \nabla u |^{p-2} \nabla u \nabla \phi\,dx=\int_{\Omega} [a(x)f(u)+\lambda b(x)g(u)+\mu V(x, \nabla u)]\phi\,dx, \] for all $\phi \in C_0^\infty(\Omega)$. In this article we say that a function $h:(0, \infty)\to [0, \infty)$ is $(p-1)$-sublinear at $0$ or at $+\infty$, if $\lim_{s\to 0} h(s)/s^{p-1}=\infty$ or $\lim_{s\to \infty} h(s)/s^{p-1}=0$ respectively; $(p-1)$-superlinear at $0$ or at $+\infty$, if $\lim_{s\to 0} h(s)/s^{p-1}=0$ or $\lim_{s\to \infty} h(s)/s^{p-1}=\infty$ respectively and $(p-1)$-asymptotically linear, if there are positive and finite numbers that correspond to the values of these limits. To abbreviate, we say sublinear, superlinear and asymptotically linear nonlinearities, respectively. In particular, a sublinear term $h$ at $0$ is called singular at $0$, if $ \lim_ {s \to 0 ^ +} h(s)= \infty$. For $\mu = 0$, problems like \eqref{1} have been intensively studied in recent years, including sublinear and superlinear nonlinearities terms at zero and/or infinity and singular terms at zero. In \cite{CRT,LS,HS,AhM3} and references therein studies on the bounded domain case are found. For $\Omega=\mathbb{R}^N$, we refer the reader to \cite{CRU,LS1,MO,CAS3} and their references. However, there are not many results in the case where the nonlinearities depend on the gradient of the solution, that is, $\mu\neq 0$, with $p\neq 2$. In general, variational techniques are not suitable to handle \eqref{1}. In the case $p=2$, an interesting exception can be seen in \cite{FGM}. One of the novelties in this article is that we improve a regularization-mono\-tonicity technique (see Section 2). This allows us to treat \eqref{1} with superlinear nonlinearities both in bounded domain and $\mathbb{R}^N$. This improvement also makes possible for us to study \eqref{1} in $\mathbb{R}^N$ by creating a sequence of solutions of \eqref{1} in bounded domains, locally bounded below by a positive function and bounded above by a carefully constructed function. We emphasize that our results do not require any monotonicity condition and (or) singularity of the functions $ f $ and $ g $. We are particularly interested in the cases where $ f$ and $ g $ may have singularity at $0$. Problems including singular nonlinearities arise in electrical conductivity, the theory of pseudoplastic fluids, singular minimal surfaces, reaction-diffusion processes, the obtaining of various geophysical indexes and industrial processes, among others; see \cite{CN,FM} for a detailed discussion. Considering the problem \eqref{1} in smooth bounded domains, we quote Zhang and Yu \cite{ZY} who, in 2000, studied the problem \begin{equation} \label{2} \begin{gathered} - \Delta u = u^{-\alpha} +\lambda +\mu |\nabla u|^q\quad \text{in } \Omega,\\ u > 0\quad \text{in }\Omega,\quad u=0 \quad \text{on } \partial\Omega, \end{gathered} \end{equation} where $\mu, \lambda\geq 0$, $\alpha>0$ and $q \in (0, 2]$. Using a change of variables, the authors proved that the problem \eqref{2} has classical solutions for $\mu\lambda<\lambda_{1}$, if $q=2$ or $\mu \in [0, {\mu^*})$, if $00$ denotes the first eigenvalue of the Dirichlet problem in $W^{1,2}_0(\Omega)$. Ghergu and Radulescu \cite{GR3} considered \begin{equation} \label{3} \begin{gathered} - \Delta u = h(u) +\lambda f(x, u)+\mu|\nabla u |^q \quad \text{in } \Omega,\\ u>0 \quad \text{in } \Omega, \quad u= 0 \quad \text{on } \partial\Omega, \end{gathered} \end{equation} under the conditions $f>0$ in $\overline{\Omega}\times (0, \infty)$, $\partial f/\partial s (x, s)\geq 0$, $s>0$, $f(x, s)/{s}$ non-increasing in $s>0$, $\lim_{s\to\infty}f(x,s)/{s}=0$, $\lim_{s\to 0}h(s)=+\infty$, $h \in C^{0, \alpha}((0, \infty))$, $h>0$ non-increasing and $\lambda=1$. They proved that \begin{itemize} \item[(i)] if $00$ such that \eqref{3} has a solution for $0\leq\mu<\mu^*$. Moreover, if $10 \quad \text{in }\Omega ,\quad u= 0 \quad \text{on } \partial\Omega, \end{gathered} \end{equation} where $g$ and $V$ are locally H\"older continuous functions such that $$ b |s|^{r_1}\leq g(x, s)\leq a_1(x)+a_2(x)| s |^{r_2}+\frac{a_3(x)}{|s |^{r_3}},\quad 0\leq V(x, \xi)\leq a_5(x)+a_4(x)|\xi|^{r_4}, $$ with $b>0 , r_i \in (0, 1)$, $i=1,\dots ,4$ are constants and $a_i$, $i=1,\dots ,5$ are positive continuous functions. Under these conditions, it was shown that \eqref{4} has a solution for each $\mu\geq 0$. Liu, Shi and Wei \cite{LSW}, still with $p=2$, recently showed, by using Morse theory and an iterative method, existence of solution for a problem like \eqref{1} with terms that have asymptotically linear growth at zero and infinity. Considering singular terms at $0$ and permitting $p\neq 2$, Loc and Schmitt \cite{LSc} used the lower and upper solution method to show existence of solution for \eqref{1} with the nonlinearity of the gradient term bounded above by the natural growth. For $\Omega=\mathbb{R}^N$ in the problem \eqref{1}. In 2007, Ghergu and Radulescu \cite{GR1} showed existence of solution for the problem \begin{equation} \label{5} \begin{gathered} - \Delta u = a(x)[f(u)+g(u)+| \nabla u |^q] \quad\text{in } \mathbb{R}^N,\\ u > 0 \quad \text{in } \mathbb{R}^N,\quad\text{and}\quad u(x)\buildrel |x| \to \infty \over \longrightarrow 0, \end{gathered} \end{equation} where $q \in (0, 1)$, $f \in C^1((0, \infty))$ is positive and decreasing, $\lim_{s\to 0^+} f(s)=\infty$ and the function $g: [0, \infty)\to [0, \infty)$ satisfies \begin{gather*} g'\geq 0, \quad \frac{g(s)}{s} \text{ is non-increasing in } s>0,\quad \lim_{s\to 0^+}\frac{g(s)}{s}=+\infty \\ \text{and } \lim_{s\to \infty}\frac{g(s)}{s}=0. \end{gather*} Concerning the function $a$, they assumed that $00$ the first eigenvalue and by $\varphi_\Omega=\varphi_{1,\Omega}>0$ the first eigenfunction of the problem \begin{equation}\label{7} \begin{gathered} - \Delta_p \varphi = \lambda \rho(x) |\varphi|^{p-2} \varphi \quad \text{in } \Omega,\\ \varphi > 0 \quad \text{in } \Omega, \quad \varphi=0 \quad \text{on } \partial \Omega, \end{gathered} \end{equation} where $\Omega \subset \mathbb{R}^N$ is a smooth bounded domain. Moreover, we denote by $\lambda_1(\rho)=\lim_{R\to\infty} \lambda_{1, B_R(0)}(\rho)\geq0$, where $B_R(0)$ is the ball centered at the origin of $\mathbb{R}^N$ with radius $R>0$. Also, we let us assume: \begin{itemize} \item[(V1)] $ V(x, \xi)\leq \alpha(x) |\xi |^q +\beta(x)~ \text{in}~\Omega\times\mathbb{R}^N$ for some $0 \leq\alpha, \beta \in C(\Omega)\cap L^\infty(\Omega)$ and $q \geq 0$, \item[(M1)] there exists $\omega_M \in C^1(\overline{\Omega})$ ($\omega_M \in C^1({\Omega})\cap W^{1,\infty}(\Omega)$ if $\Omega=\mathbb{R}^N$) satisfying \begin{equation}\label{6} \begin{gathered} -\Delta_p \omega_M = M(x)\quad \text{in } \Omega,\\ \omega_M > 0\quad \text{in } \Omega,\quad \omega_M=0\quad \text{on } \partial\Omega, \end{gathered} \end{equation} where $M(x):=\max\{2a(x), 2b(x), \alpha(x), \beta(x)\}$, $x \in \Omega$, \item[(F1)] \[ (F_0)\quad f_0<1/ \|\omega_M\|_{L^\infty(\Omega)}^{p-1},\quad \text{or}\quad (F_{\infty})\quad f_\infty<1/\|\omega_M\|_{L^\infty(\Omega)}^{p-1}. \] \end{itemize} \begin{remark} \label{rmk1.1} {\rm With respect to the hypotheses (M1) and (F1), we note that: \begin{enumerate} \item If $\Omega\subset\mathbb{R}^N$ is a smooth bounded domain, then (M1) occurs if, for example, $M \in L^q(\Omega)$ for some $q>N>1$. See, for instance, Perera and Zhang \cite{PZ1}. This allows we take singular potentials of the type $a(x)=b(x)={1}/{(1-|x|)^\gamma}$, with $\gamma<1$ and $\Omega=B_1(0)\subset\mathbb{R}^N$ in \eqref{1}. \item If $\Omega=\mathbb{R}^N$, it is known that \eqref{6} has a solution if $M$ is a bounded continuous function and satisfies $$ M_\infty:=\int_0^\infty\Big[s^{1-N}\int_0^s t^{N-1}\hat{M}(t)dt\Big]^{\frac{1}{p-1}}ds<\infty, $$ where $\hat{M}(t)=\max_{| x |=t} M(x)$, $t\geq 0$. The existence and $L^{\infty}$-boundedness of a solution of \eqref{6} imply its regularity (see \cite{DiB}). In addition, if we assume that $N\geq 3$ and $$ \int_1^\infty r^{\frac{1}{p-1}}\hat{M}^{\frac{1}{p-1}}(r) dr<\infty \quad \text{or}\quad \int_1^\infty r^{\frac{(p-2)N+1}{p-1}}\hat{M}(r)dr<\infty, $$ if $1\lambda_\Omega(\rho),\\ \max\{0,\frac{\lambda_{\Omega}(\rho)-f_0}{g_0}\}, &\text{if } 0 0$ and a $ u=u_{\lambda, \mu} \in C^1(\Omega)\cap C(\overline{\Omega})$ solution of \eqref{1} for each $0\leq\mu<\mu^*$. Additionally: \begin{itemize} \item[(i)] $u\geq c\varphi_\Omega$ for some $c>0$, \item[(ii)] if $(F_i)$ holds, for $i\in \{0,\infty\}$, then $$ \lambda^*\geq \frac{1}{g_i}\Big(\frac{1}{\|\omega_M\|_{L^\infty(\Omega)}^{p-1}}-f_i\Big) :=\lambda^i, $$ \item[(iii)] there exists a constant $d>0$ such that $$ \mu^*_\lambda\geq d\min\big\{{[f^i+\lambda g^i]^{\frac{p-1-q}{p-1}}}, {f^i+\lambda g^i}\big\},\quad \text{if } q\in [0,p-1]. $$ \end{itemize} \end{theorem} For $\Omega=\mathbb{R}^N$ and $1 0$ and a $ u=u_{\lambda, \mu} \in C^1(\mathbb{R}^N)$ solution of \eqref{1} for each $0\leq\mu<\mu^*$. Moreover, if $(F_i)$ holds, for $i \in \{0,\infty\}$, then there is a constant $d>0$ such that \begin{itemize} \item[(i)] $\lambda^*\geq \lambda^i$ \item[(ii)] $\mu^*_\lambda\geq d\min\big\{{[f^i+\lambda g^i]^{\frac{p-1-q}{p-1}}}, {f^i+\lambda g^i}\big\}$ for $0<\lambda<\lambda^i$. \end{itemize} \end{theorem} \begin{remark} \label{rmk1.2} \rm In the definition of $\lambda_{*}$, the possibility $f_0>\lambda_{1}(\rho)$ does not permit $(F_0)$ to occur, because $\lambda_\Omega(\rho)\geq \lambda_{\Omega}(M)\geq \Vert w_M\Vert^{1-p}_{L^{\infty}(\Omega)}$ and as a consequence of this, we have $\lambda_{1}(\rho)\geq \lambda_{1}(M)\geq \Vert w_M\Vert^{1-p}_{L^{\infty}(\mathbb{R}^N)}$ also (see Santos \cite{CAS3}). In this situation, $(F_{\infty})$ should occur, as in \cite{GR1} and \cite{XZ}. \end{remark} Theorem \ref{NL+} improves previous results principally because it addresses the p-Laplacian operator, obtains estimates for $\lambda^{*} $ and $\mu^{*} $, no monotonicity or growth restriction on the nonlinearities are required, the cases $q=0$ and $q=p-1 $ are included and we assume the hypothesis (M1) that is weaker than \eqref{5a}. We point out that problem \eqref{1} has no solution for $p\geq N$ (see Serrin and Zou \cite{sz}). This paper is organized as follows: In section 2 we construct several auxiliary functions for the terms $f$ and $g$ and we study their properties. Because of the singularities allowed on $f$ and $g$, we regularize the problem \eqref{1} and we obtain an upper solution for it in bounded domain and in $\mathbb{R}^N$, in sections 3 and 5, respectively. After that, we use section 4 to prove Theorem \ref{DL+}. In section 6, we generalize this result for $\mathbb{R}^N$. \section{Auxiliary functions} To prove Theorems \ref{DL+} and \ref{NL+} we refine a regularization-motonicity technique used, among others, by Feng and Liu \cite{FL}, Zhang \cite{Z} and Mohammed \cite{MO}. Observing that we do not assume monotonicity on the nonlinearities, we introduce a truncation of the terms $f$ and $g$ through a parameter $\gamma>0$ and build auxiliary functions which allow us to obtain not only the monotonicity, but also the necessary regularity for the proof of our results. Parallel to this, the inclusion of a parameter $\theta<1$, in this construction, makes it possible solving the problem \eqref{1} for the case $q>p-1$. Analyzing the behavior of these auxiliary functions, the parameters $\lambda, \gamma, \theta$ and the fact that the problem \eqref{6} has a solution, we determine a $\Lambda^*$-curve whose behavior allow us to find region of variation for the parameter $\lambda$, and consequently, obtain an estimate from below for that region. With these purposes, let us define the continuous functions, depending on real parameter $\gamma>0$, as $$ f_ \gamma(s):= \begin{cases} f(s), &\text{if } 0 0$, defining the function \begin{equation} \label{12} {\zeta}_{\lambda, \gamma}(s)=s^{p-1}\sup \big\{\frac{{f_ \gamma}(t)}{t^{p-1}}, \;t>s\big\}+\lambda s^{p-1} \sup \big\{\frac{{g_ \gamma}(t)}{t^{p-1}}, \; t>s\big\}, \quad \lambda\geq 0 \end{equation} we obtain, from the above definitions, that \begin{itemize} \item[(i)] $\frac{{\zeta}_{\lambda, \gamma}(s)}{s^{p-1}}$ is non-increasing in $s>0$; \item[(ii)] ${\zeta}_{\lambda, \gamma}(s)\geq {f_\gamma}(s)+\lambda {g_ \gamma}(s)$, $s>0$; \item[(iii)] $\lim_{s\to\infty}\frac{{\zeta}_{\lambda, \gamma}(s)}{s^{p-1}} =\frac{f(\gamma)}{\gamma^{p-1}}+\lambda \frac{g(\gamma)}{\gamma^{p-1}}$. \end{itemize} Now, defining $$ H_{\lambda, \gamma}(s)=\frac{s^2}{\int_0^s\frac{t}{{\zeta}_{\lambda, \gamma}(t)^{\frac{1}{p-1}}}dt}, \ s>0, $$ and using (i) and (iii) above, we have the following lemma. \begin{lemma}\label{H} The function $H$ satisfies: \begin{itemize} \item[(i)] $H_{\lambda, \gamma} \in C^1((0,\infty), (0,\infty))$; \item[(ii)] ${\zeta}_{\lambda, \gamma}(s) \leq [H_{\lambda, \gamma}(s)]^{{p-1}}$, $s>0$; \item[(iii)] $\frac{H_{\lambda, \gamma}(s)}{s}$ is non-increasing in $s>0$; \item[(iv)] \[ \lim_{s\to\infty}\frac{H_{\lambda, \gamma}(s)}{s} =\Big[\frac{f(\gamma)}{\gamma^{p-1}}+\lambda \frac{g(\gamma)}{\gamma^{p-1}}\Big]^{\frac{1}{p-1}}. \] \end{itemize} \end{lemma} After these, introducing a parameter $ \theta \in (0, 1]$ and defining the function \begin{equation}\label{13} \Gamma_{\lambda}(\gamma)=\Gamma_{\lambda,\theta}(\gamma)=\frac{\theta}{\gamma}\int_0^\gamma\frac{t^\theta}{H_{\lambda, \gamma}(t^\theta)}dt, \quad \gamma>0 \end{equation} we obtain, from the previously defined functions and their properties, the following result. \begin{lemma}\label{Gamma2} Suppose {\rm (M1)} and {\rm (F1)} hold. Then for each $\theta \in (\Vert w_M \Vert_{\infty}f_i^{1/(p-1)} ,1]$, for either $i=0$ or $i=\infty$, we have: \begin{itemize} \item[(i)] $\lim_{\gamma\to\infty}\Gamma_{\lambda, \theta}(\gamma) =\frac{\theta}{(f_\infty+\lambda g_\infty)^{\frac{1}{p-1}}}$, for each $\lambda\geq 0$; \item[(ii)] $\lim_{\gamma\to 0}\Gamma_{\lambda, \theta}(\gamma)=\frac{\theta}{(f_0+\lambda g_0)^{\frac{1}{p-1}}}$, for each $\lambda\geq 0$; \item[(iii)] $\Gamma_{\lambda, \theta}$ is decreasing in $\lambda>0$, for each $\gamma>0$; \item[(iv) ] there exists a $\tilde{\gamma}=\tilde{\gamma}(\Omega, \theta)>0$ such that $\Gamma_{0,\theta}(\tilde{\gamma})>\|\omega_M\|_{L^\infty(\Omega)}$. \end{itemize} \end{lemma} By Lemma \ref{Gamma2}, we can define the nonempty set $$ \mathcal{A}=\mathcal{A}_{\Omega, \theta}:=\{{\gamma} \in (0, \infty) : \Gamma_{0,\theta}({\gamma})>\|\omega_M\|_{L^\infty(\Omega)}\}. $$ Now, as a consequence of $\lim_{\lambda\to \infty}\Gamma_{\lambda,\theta}(\gamma)=0$, $\lim_{\lambda\to 0}\Gamma_{\lambda,\theta}(\gamma)=\Gamma_{0,\theta}(\gamma)$ and of the above lemma, we have that the function $\Lambda^*=\Lambda^*_{\Omega, \theta}:\mathcal{A}\to (0, \infty)$ that associate for each ${\gamma} \in \mathcal{A}$ the unique number $\Lambda^*({\gamma})$ satisfying \begin{equation}\label{14} \Gamma_{\Lambda^*({\gamma}),\theta}({\gamma})=\|\omega_M\|_{L^{\infty}(\Omega)}, \end{equation} is well defined. Thus, we can define the positive number \begin{equation}\label{lam} \lambda^*_{\theta}({\Omega}):=\sup\{\Lambda^*({\gamma}): {\gamma} \in \mathcal{A}\}. \end{equation} After these, we infer the following lemma. \begin{lemma}\label{eta2} Suppose {\rm (M1)} and {\rm (F1)} hold. Then for each $\theta \in (\Vert w_M \Vert_{\infty}f_i^{1/(p-1)},1]$, we have $$ \lambda^*_\theta(\Omega)\geq \frac{1}{g_i} \Big(\frac{\theta}{\|\omega_M\|_{L^\infty(\Omega)}^{p-1}}-f_i\Big) :=\lambda^i_{\theta}. $$ \end{lemma} \begin{proof} If $(F_0)$ occurs and $g_0<\infty$, then for each $0<\delta<{\lambda}^0_{\theta}$, from Lemma \ref{Gamma2} (ii) it follows that \begin{align*} \liminf_{\gamma\to 0} (\Gamma_{\delta,\theta}(\gamma)-\|\omega_M\|_\infty) &=\frac{\theta}{(f_0+\delta g_0)^{\frac{1}{p-1}}}-\|\omega_M\|_\infty\\ &>\frac{\theta}{(f_0+{\lambda}^0_{\theta} g_0)^{\frac{1}{p-1}}}-\|\omega_M\|_\infty= 0. \end{align*} Now, if $(F_{\infty})$ occurs and $g_{\infty}<\infty$, using Lemma \ref{Gamma2} (i), we have \begin{align*} \liminf_{\gamma\to \infty} (\Gamma_{\delta,\theta}(\gamma)-\|\omega_M\|_\infty) &=\frac{\theta}{(f_\infty+\delta g_\infty)^{\frac{1}{p-1}}}-\|\omega_M\|_\infty\\ &>\frac{\theta}{(f_\infty+{\lambda^{\infty}_{\theta}} g_\infty)^{\frac{1}{p-1}}}-\|\omega_M\|_\infty=0, \end{align*} for each $0<\delta<{\lambda^{\infty}_{\theta}}$. So, in both cases, there exists a $\gamma_0=\gamma_0(\delta)>0$ such that $\Gamma_{\delta,\theta}(\gamma_0)>\|\omega_M\|_\infty$. As a consequence of this and Lemma \ref{eta2}(iii), we have that $\gamma_0 \in \mathcal{A}$, because $\Gamma_{0, \theta}(\gamma_0)>\Gamma_{\delta, \theta}(\gamma_0)>\|\omega_M\|_\infty$. So, from \eqref{14} there is a unique $\Lambda^*(\gamma_0)$ such that $\Gamma_{\Lambda^*(\gamma_0),\theta}(\gamma_0)=\|\omega_M\|_\infty$. Now, using $\Gamma_{\Lambda^*(\gamma_0),\theta}(\gamma_0)<\Gamma_{\delta,\theta}(\gamma_0)$ and Lemma \ref{eta2}(iii), we obtain $\Lambda^*(\gamma_0)>\delta$. So, by the arbitrariness of $\delta$, it follows the proof of the Lemma. \end{proof} Now, defining \begin{equation}\label{15} \eta_\lambda(s)=\eta_{\lambda,\theta}(s) =\frac{\theta}{{\gamma}}\int_0^s\frac{t^{\theta}}{H_{\lambda, {\gamma}}(t^{\theta})}dt, \quad s>0, \; {\gamma} \in \mathcal{A}, \; \lambda>0, \end{equation} it follows that \begin{equation}\label{a+} \eta_{\lambda,\theta}({\gamma})=\Gamma_{\lambda,\theta}({\gamma}) >\|\omega_M\|_\infty+\bar{\sigma}, \end{equation} for each $0<\lambda<\Lambda^*({\gamma})$, where $\bar{\sigma}=\bar{\sigma}(\lambda, \theta, {\gamma})=\big(\Gamma_{\lambda,\theta}({\gamma})-\|\omega_M\|_\infty\big)/2>0$. Besides this, the following lemma follows from the previous results. \begin{lemma}\label{eta} Suppose {\rm (M1)} and {\rm(F1)} hold. Then, for each $0<\lambda<\lambda^*_\theta(\Omega)$ given: \begin{itemize} \item[(i)] $[\bar{\sigma}, \|\omega_M\|_\infty+\bar{\sigma}]\subset \operatorname{Im}(\eta_\lambda)$; \item[(ii)] $\eta_\lambda \in C^2((0, \infty), \operatorname{Im}(\eta_\lambda))$ is increasing in $s>0$; \item[(iii)] $ \eta_\lambda^{-1}:=\psi_\lambda \in \ C^2((\operatorname{Im} (\eta_\lambda)\backslash \{0\}, (0,\infty))$ is increasing in $s>0$; \item[(iv)] $\psi_\lambda'(s)=\frac{{\gamma} H_{\lambda, \gamma_0}(\psi_\lambda(s)^{\theta})}{\theta\psi_\lambda(s)^{\theta}}$, $s>0$; \item[(v)] $\psi_\lambda''(s)\leq 0$, $s>0$; \item[(vi)] $\eta_\lambda$ is decreasing in $\lambda$. \end{itemize} \end{lemma} \section{An auxiliary problem} To solve the problem \eqref{1} with the gradient term in the presence of nonlinearities $f$ and $g$ already described, we will explore the behavior of the auxiliary $\lambda, \gamma, \theta$-functions given in the previous section considering different intervals of variation for $q\in [0, p]$ and an appropriate division of the domain $\Omega\subset\mathbb{R}^N$. All this together with the behavior of the $\Lambda^*$-curve will allow us to determine a $\mu^*$-curve whose behavior will define the region of variation of the parameter $\mu\geq 0$. As a consequence of the hypotheses (M1), (F1) and of the behavior of $\Lambda^*, \mu^* $-curves, we obtain a $\gamma_0$ which allow us to show the existence of solution ($\epsilon$-uniformly limited in $ L^\infty(\Omega)$) of the $\epsilon $-family of problems \eqref{8} below, for appropriate $ \lambda>0$ and $\mu \geq 0$. In this sense, we will construct a positive bounded upper solution for the $\epsilon$-family of problems \begin{equation} \label{8} \begin{gathered} - \Delta_p u = a(x)f(u+\epsilon) + \lambda b(x)g(u+\epsilon) +\mu V(x, \nabla u) \quad \text{in } \Omega\\ u>0 \quad\text{in } \Omega,\quad u=0 \quad \text{on } \partial \Omega, \end{gathered} \end{equation} for sufficiently small $\epsilon>0$. \begin{proposition}\label{SuperDL} Assume {\rm (F1), (M1), (V1)} with $q \in [0, p]$ hold. Then there exists a $ \lambda^*>0$ such that for each $0<\lambda<\lambda^*$, there exist real numbers $\overline{\sigma}=\overline{\sigma}(\lambda)>0$ and $\mu^*=\mu^*_\lambda>0$, both independent of $\epsilon$, such that if $0<\sigma\leq\overline{\sigma}$ and $0\leq\mu<\mu^*$, then there exists a $ v_\sigma=v_{\sigma, \lambda} \in C^1(\overline{\Omega})$ upper solution of $\eqref{8}$. Additionally: \begin{itemize} \item [(i)] $\psi_\lambda({\sigma})^{\theta_0} \leq v_{{{\sigma}}}\leq \gamma_0^{\theta_0}$ for some $\theta_0=\theta_0(\lambda) \in (\Vert w_M \Vert_{\infty}f_i^{1/(p-1)} ,1]$ and $\gamma_0=\gamma_0(\lambda)>0$; \item [(ii)] if $(F_ i)$ holds, for $i \in \{0,\infty\}$, then $$ \lambda^*\geq \frac{1}{g_i}\Big(\frac{1}{\|\omega_M\|_{L^\infty(\Omega)}^{p-1}}-f_i\Big) :=\lambda^i; $$ \item [(iii)] there exists a constant $d>0$ such that for $0<\lambda<\lambda^i$, we have $$ \mu^*_\lambda\geq d\min\big\{{[f^i+\lambda g^i]^{\frac{p-1-q}{p-1}}}, {f^i+\lambda g^i}\big\} \quad \text{if } q \in [0,p-1]. $$ \end{itemize} \end{proposition} \begin{proof} Because the possible singular behavior of the nonlinearities, we divide this proof into two parts, depending on the value of the exponent $q$ of the gradient term in the hypothesis (V1). \smallskip \noindent\textbf{Case one:} $q \in [0, p-1]$. In this case, we pick $\theta_0=1$ and take $\theta=\theta_0$ in the functions $\Gamma_{\lambda, \theta}$ and $\eta_{\lambda,\theta}$. So, given $0<\lambda<\lambda^*:=\lambda^*_1(\Omega)$ we define, for each ${\gamma}>0$, the positive number \begin{equation}\label{ex} \mu^*_{\lambda}({\gamma})=\mu^*_{\lambda, \Omega}({\gamma}):=\min\Big\{\frac{[{f}({\gamma})+\lambda {g}({\gamma})]^{\frac{p-1-q}{p-1}}}{4\|\nabla\omega_M\|_{L^\infty(\Omega)}^q}, \frac{{f}({\gamma})+\lambda {g}({\gamma})}{4}\Big\}. \end{equation} Now, we can define \begin{equation}\label{ex1} \mu^*_\lambda=\mu^*_{\lambda, \Omega}:=\sup\{\mu^*_{\lambda}({\gamma}): {\gamma} \in \mathcal{A}\ \text{and} \ \lambda<\Lambda^*({\gamma})\}\in (0,\infty]. \end{equation} So, from \eqref{lam}, there exists $\overline{\gamma} \in \mathcal{A}$ such that $\lambda<\Lambda^*(\overline{\gamma})$. That is, $\mu^*_\lambda\geq\mu^*_{\lambda}(\overline{\gamma})>0$. Thus, given $0\leq\mu<\mu^*_\lambda$ there is a $\gamma_0 = \gamma_0(\lambda)\in \mathcal{A}$ such that $\lambda < \Lambda^*(\gamma_0)$ and $\mu<\mu^*_{\lambda}(\gamma_0)$. Now, we fix this $\gamma_0$. From the hypothesis (M1) and Lemma \ref{eta} (ii), we define $v_{\sigma}=v_{\sigma,\lambda} \in C^1(\overline{\Omega})$, increasing in $\sigma$, by \begin{equation}\label{17} v_{\sigma}(x):=\psi_\lambda(\omega_M(x)+{\sigma}), \ x \in \overline{\Omega} \end{equation} for each $0<\sigma \leq \bar{\sigma}$, where $\bar{\sigma}=\bar{\sigma}(\lambda)$ is given in \eqref{a+}. So, $v_\sigma(x)>\psi_\lambda(\sigma)$ in $\Omega$ and $v_\sigma(x)=\psi_\lambda({\sigma})$ on $\partial \Omega$, because $\omega_M(x)>0$ in $\Omega$ and $\omega_M(x)=0$ on $\partial\Omega$. Besides this, from \eqref{a+}, Lemma \ref{eta} (iii) and $0<\lambda<\Lambda^*(\gamma_0)$ we have that $v_{\overline{\sigma}}(x)<\gamma_0, \ x \in \overline{\Omega}$. So, there exists an $\epsilon>0$, which is sufficiently small, such that \begin{equation}\label{18} \| v_{\sigma}\|_{L^\infty(\Omega)}<\gamma_0-\epsilon,~0<\sigma \leq \bar{\sigma}. \end{equation} Now, it follows from \eqref{17}, Lemmas \ref{H}, \ref{eta} and the assumption (M1), that \begin{equation}\label{19} \begin{aligned} &\int_{\Omega}|\nabla v_{\sigma}|^{p-2}\nabla v_{\sigma} \nabla\phi\, dx\\ &=\int_{\Omega}[\psi_\lambda'(\omega_M)+\sigma]^{p-1} |\nabla\omega_M|^{p-2} \nabla\omega_M\nabla\phi\,dx\\ &= \int_{\Omega}|\nabla\omega_M|^{p-2} \nabla\omega_M \nabla([\psi_\lambda'(\omega_M+\sigma)]^{p-1} \phi)\, dx\\ &\quad -(p-1)\int_{\Omega}|\nabla\omega_M|^p [\psi_\lambda'(\omega_M+\sigma)]^{p-2}\psi_\lambda''(\omega_M+\sigma)\phi\,dx\\ &\geq \int_{\Omega} M(x)[\psi_\lambda'(\omega_M+\sigma)]^{p-1} \phi\,dx\\ &=\int_{\Omega} M(x)\gamma_0^{p-1}\Big[\frac{H_{\lambda, \gamma_0}(\psi_\lambda(\omega_M+\sigma))}{\psi_\lambda(\omega_M+\sigma)}\Big]^{p-1} \phi\, dx \end{aligned} \end{equation} for each $\phi \in C_0^\infty(\Omega)$, $\phi\geq 0$. The study of this inequality will be divided in two parts. One of them will produce an estimate for $af+\lambda bg$ while the other will result in an estimate for $\mu V$. We note that from the definitions and properties of the functions defined in the Section 2 and \eqref{18} that \begin{equation}\label{20} \begin{aligned} &\frac{1}{2}\int_{\Omega} M(x)\gamma_0^{p-1}\Big[\frac{ H_{\lambda, \gamma_0}(\psi_\lambda(\omega_M+{\sigma}))}{\psi_\lambda(\omega_M+{\sigma})} \Big]^{p-1}\phi \\ &\geq\frac{1}{2}\int_{\Omega} M(x)\gamma_0^{p-1}\frac{{\zeta}_{\lambda, \gamma_0}(v_{\sigma}+\epsilon)}{(v_{\sigma}+\epsilon)^{p-1}} \phi \\ &\geq \frac{1}{2}\int_{\Omega} M(x)\gamma_0^{p-1}\frac{{\zeta}_{\lambda, \gamma_0}(v_{\sigma}+\epsilon)}{(\gamma_0)^{p-1}} \phi\\ &\geq\int_{\Omega} [a(x) f(v_{\sigma}+\epsilon)+\lambda b(x)g(v_{\sigma}+\epsilon)]\phi \end{aligned} \end{equation} for each $\epsilon>0$ and $0<\sigma<\bar{\sigma}$. On the other hand, from Lemma \ref{H} (iii)-(iv) and $0 \leq q \leq p-1$, it follows that \begin{align*} &\frac{1}{2}\int_{\Omega} M(x)\gamma_0^{p-1}\Big[\frac{ H_{\lambda, \gamma_0}(\psi_\lambda(\omega_M+{\sigma}))}{\psi_\lambda(\omega_M+{\sigma})}\Big]^{p-1}\phi\,dx\\ &\geq \frac{1}{4}\int_{\Omega} M(x)\gamma_0^{p-1}\Big[\frac{f(\gamma_0)}{\gamma_0^{p-1}}+\lambda \frac{g(\gamma_0)}{\gamma_0^{p-1}}\Big] \phi\,dx \\ &\quad +\frac{1}{4}\int_{\Omega} M(x)\gamma_0^{p-1-q}\Big[\frac{ H_{\lambda, \gamma_0}(\psi_\lambda(\omega_M+{\sigma}))}{\psi_\lambda(\omega_M+{\sigma})} \Big]^{p-1-q}\Big[\frac{ \gamma_0 H_{\lambda, \gamma_0}(\psi_\lambda(\omega_M+{\sigma}))} {\psi_\lambda(\omega_M+{\sigma})}\Big]^q\phi\,dx\\ &\geq \frac{[{f(\gamma_0)}+\lambda {g(\gamma_0)}]}{4}\int_\Omega M(x)\phi\,dx\\ &\quad + \frac{\{[{f(\gamma_0)}+\lambda {g(\gamma_0)}]^{\frac{1}{p-1}}\}^{p-1-q}}{4}\int_\Omega M(x)[\psi'_\lambda(\omega_M+{\sigma})]^q \phi\,dx\\ &\geq \frac{[f(\gamma_0)+\lambda {g(\gamma_0)}]}{4}\int_\Omega \beta(x)\phi\,dx\\ &\quad +\frac{[{f}({\gamma_0})+ \lambda {g}({\gamma_0})]^{\frac{p-1-q}{p-1}}}{4\|\nabla\omega_M\|_{L^\infty(\Omega)}^q} \int_\Omega M(x)[\psi'_\lambda(\omega_M+{\sigma})]^q |\nabla\omega_M |^q \phi\,dx. \end{align*} Now, using (V1) and \eqref{ex} we can write \begin{equation}\label{5+} \begin{aligned} &\frac{1}{2}\int_{\Omega} M(x)\gamma_0^{p-1}\Big[\frac{ H_{\lambda, \gamma_0}(\psi_\lambda(\omega_M+{\sigma}))}{\psi_\lambda(\omega_M+{\sigma})} \Big]^{p-1}\phi\,dx\\ &\geq \mu^*_{\lambda}(\gamma_0)\int_\Omega\beta(x)\phi\,dx +\mu^*_{\lambda}(\gamma_0)\int_\Omega\alpha(x)| \nabla v_{\sigma} |^q \phi\,dx \geq \mu\int_\Omega V(x, \nabla v_{\sigma})\phi\,dx. \end{aligned} \end{equation} So, replacing \eqref{20} and \eqref{5+} in \eqref{19}, we conclude the proof of Proposition \ref{SuperDL}. \smallskip \noindent\textbf{Case two:} $q \in (p-1, p]$. If $g_i<\infty$, we define $\lambda^*:=\liminf_{\theta\nearrow 1}\lambda^*_\theta(\Omega)$, where $\lambda^*_{\theta}(\Omega)$ was defined in \eqref{lam}. Note that, by Lemma \ref{eta2}, we have $\lambda^*\geq \lambda^i_{\theta}$ with $\theta=1$. So, given $0<\lambda<\lambda^*$, there is a $\theta_0=\theta_0(\lambda) \in (\Vert w_M \Vert_{\infty}f_i^{1/(p-1)} ,1)$ such that $0<\lambda<\lambda^*_{\theta_0}(\Omega).$ Now, we fix this $\theta_0$ in the functions $\Gamma_{\lambda, \theta}$ and $\eta_{\lambda, \theta}$ defined in \eqref{13} and \eqref{15}, respectively. So, if $g_i=\infty$, we choose a $\theta_0 \in (\Vert w_M \Vert_{\infty}f_i^{1/(p-1)} ,1)$ and we set $\lambda^*=\lambda^*_{\theta_0}(\Omega)$. In this case, we have $\lambda^*\geq \lambda^i_{\theta_0}$. In both cases, given $0<\lambda<\lambda^*$, we set the positive number $\mu^*_\lambda(\gamma):=\mu^*_{\lambda, \Omega}({\gamma})$ by \begin{equation}\label{mu'} \begin{aligned} \min\Big\{&\frac{\gamma^{p-1-q}}{4C_2\|\nabla\omega_M\|^q_{L^\infty(\Omega)}} \frac{\gamma^{(p-1)(\theta_0-1)}[f(\gamma)+\lambda g(\gamma)]}{4},\\ &\frac{(1-\theta_0)(p-1)[\gamma H_{\lambda, \gamma}(1)]^{p-q}}{4\|\alpha\|_{L^\infty(\Omega)}}\Big\} \end{aligned} \end{equation} for each $\gamma>0$ and for some constant $C_2=C_2(\gamma)>0$ to be chosen posteriorly. Now, we define $$ \mu^*_\lambda=\mu^*_{\lambda, \Omega}:=\sup\big\{\mu^*_\lambda(\gamma): \gamma \in \mathcal{A} \ \text{and} \ \lambda<\Lambda^*(\gamma)\big\}. $$ As in Case one, we claim that $\mu^*_\lambda>0$ and given $0\leq\mu<\mu^*_\lambda$, there is a $\gamma_0=\gamma_0(\lambda) \in \mathcal{A}$ such that $\lambda<\Lambda^*(\gamma_0)$ and $\mu<\mu^*_\lambda(\gamma_0)$. From now on, we fix this $\gamma_0$. Since $\omega_M \in C^1(\overline{\Omega}) $ and ${\partial\omega_M}/{\partial\nu}<0 $ on $\partial\Omega$, there are $\delta_0>0$ sufficiently small and $k_0=k_0(\delta_0)>0$ such that \begin{equation}\label{11+} |\nabla\omega_M|^p>{k_0(\delta_0)} \quad \text{for } x \in \Omega_{\delta_0}, \end{equation} where $\Omega_{\delta_0}=\{ x \in \Omega: \text{dist}(x, \partial\Omega)<\delta_0\}$ and $\nu$ is the exterior normal to the $\partial\Omega$. In a similar way to that done in \eqref{17}, we obtain that \begin{equation}\label{defv} v_{{\sigma}}(x):=[\psi_\lambda(\omega_M(x)+{\sigma})]^{\theta_0}, \quad x \in \overline{\Omega} \end{equation} is well-defined, $\psi_\lambda(\sigma)^{\theta_0} \leq v_\sigma\in C^1(\overline{\Omega})$ and $\|v_\sigma\|_{L^\infty(\Omega)}<\gamma_0^{\theta_0}$, for each $ 0<\sigma\leq\bar{\sigma}$. In the last conclusion, we used Lemma \ref{eta} and the inequality \eqref{a+}. That is, there is a sufficiently small $\epsilon>0$ such that \begin{equation}\label{10+} \| v_\sigma\|_{L^\infty(\Omega)}<\gamma_0^{\theta_0}-\epsilon. \end{equation} Since $\lim_{s\to 0} \psi_\lambda(s)=0$, we can take $0<\tilde{\sigma}<\bar{\sigma}$ sufficiently small such that \begin{equation}\label{d} \psi_\lambda(\tilde{\sigma})^{\theta_0}<\frac{1}{2} \quad\text{and} \quad \frac{k_0(\delta_0)}{2\psi_\lambda(\tilde{\sigma})^{\theta_0}} >\|\nabla\omega_M\|^q_{L^\infty(\Omega)}. \end{equation} So, from Lemma \ref{eta} (iii), it follows that $v_\sigma(x)0$ sufficiently small such that \begin{equation}\label{10+"} v_{{\sigma}}(x)< v_{\tilde{\sigma}}(x)< 2\psi_\lambda({{\tilde{\sigma}}})^{\theta_0}, \quad \text{for } x \in \Omega_{\delta_1}, \; \sigma \in (0, \tilde{\sigma}). \end{equation} Then, from \eqref{11+}, \eqref{d} and \eqref{10+"}, we have \begin{equation}\label{1.27} \frac{|\nabla\omega_M(x)|^p}{v_{\sigma}(x)}>\frac{k_0(\delta_0)}{2 \psi_\lambda(\tilde{\sigma})^{\theta_0}}> |\nabla\omega_M(x)|^q, \end{equation} for each $ x \in \Omega_\delta$, where $\delta=\min\{\delta_0, \delta_1\}>0$. Now, given $\phi \in C_0^\infty(\Omega)$ with $\phi\geq 0$ and $0<\sigma<{\tilde{\sigma}}$, we take $ \tau\in C_0^\infty(\Omega)$ defined by $\tau= 1$ in $\Omega\backslash\Omega_\delta$ and $\tau= 0$ in $\Omega_{\delta/2}$ with $0\leq\tau\leq 1$. So, writing $\phi = \tau \phi + (1-\tau) \phi$, we have that \begin{equation}\label{15+} \int_\Omega| \nabla v_{\sigma}|^{p-2}\nabla v_{\sigma} \nabla \phi =\int_{\Omega\backslash\Omega_{\delta/2}}|\nabla v_{\sigma} |^{p-2}\nabla v_{\sigma} \nabla (\tau\phi) +\int_{\Omega_\delta}|\nabla v_{\sigma} |^{p-2}\nabla v_{\sigma} \nabla (1-\tau)\phi . \end{equation} In $\Omega\backslash\Omega_{\delta/2}$, it follows from the definition of $v_\sigma$, that \begin{align*} &\int_{\Omega\backslash\Omega_{\delta/2}}|\nabla v_{\sigma} |^{p-2}\nabla v_{\sigma} \nabla (\tau\phi)\\ &=\int_{\Omega\backslash\Omega_{\delta/2}}|\nabla\omega_M|^{p-2} \nabla\omega_M\nabla\big\{ {\theta_0}^{p-1}[\psi_\lambda(\omega_M+\sigma)]^{(\theta_0-1)(p-1)} [\psi'_\lambda(\omega_M+\sigma)]^{p-1}\tau\phi\big\}\\ &\quad-(\theta_0-1)(p-1)\int_{\Omega\backslash\Omega_{\delta/2}} |\nabla\omega_M|^{p} {\theta_0}^{p-1}\\ &\quad\times [\psi_\lambda(\omega_M+\sigma)]^{(\theta_0-1)(p-1)-1} [\psi'_\lambda(\omega_M+\sigma)]^{p} \tau\phi \\ &\quad -(p-1)\int_{\Omega\backslash\Omega_{\delta/2}}|\nabla\omega_M|^{p} {\theta_0}^{p-1}[\psi_\lambda(\omega_M+\sigma)]^{(\theta_0-1)(p-1)}\\ &\quad\times [\psi'_\lambda(\omega_M+\sigma)]^{p-2} \psi''_\lambda(\omega_M+\sigma) \tau\phi \end{align*} Now, recalling that $\theta_0 \in (\Vert w_M \Vert_{\infty}f_i^{1/(p-1)} ,1)$, $ \psi^{\prime}_\lambda\geq 0$, $\psi_\lambda^{\prime\prime}\leq 0$ (see Lemma \ref{eta}) and noting that $$ {\theta_0}^{p-1}[\psi_\lambda(\omega_M+\sigma)]^{(\theta_0-1)(p-1)}[\psi'_\lambda(\omega_M+\sigma)]^{p-1}\tau\phi \in W_0^{1, p}(\Omega), $$ it follows from (M1) and Lemma \ref{eta} (iv) that \begin{equation} \label{16++} \begin{aligned} &\int_{\Omega\backslash\Omega_{\delta/2}}|\nabla v_{\sigma} |^{p-2}\nabla v_{\sigma} \nabla (\tau\phi) dx\\ &\geq \int_{\Omega\backslash\Omega_{\delta/2}}|\nabla\omega_M|^{p-2} \nabla\omega_M \nabla\big\{ {\theta_0}^{p-1}[\psi_\lambda(\omega_M+\sigma)]^{(\theta_0-1)(p-1)}\\ &\quad\times [\psi'_\lambda(\omega_M+\sigma)]^{p-1}\tau\phi\big\}\\ &\geq \int_{\Omega\backslash\Omega_{\delta/2}}M(x) {\theta_0}^{p-1}[\psi_\lambda(\omega_M+\sigma)]^{(\theta_0-1)(p-1)}[\psi'_\lambda(\omega_M+\sigma)]^{p-1}\tau\phi \\ &=\int_{\Omega\backslash\Omega_{\delta/2}} M(x) {\theta_0}^{p-1}{v}_{\sigma}^{\frac{(\theta_0-1)(p-1)}{\theta_0}} \frac{\gamma_0^{p-1}}{\theta_0^{p-1}}\Big[\frac{H_{\lambda, \gamma_0}((\psi_\lambda(\omega_M+\sigma))^{\theta_0})} {(\psi_\lambda(\omega_M+\sigma))^{\theta_0}}\Big]^{p-1}\tau\phi. \end{aligned} \end{equation} As in Case one, the analysis of this inequality will be divided in two parts. So, from the properties of auxiliary functions, Lemma \ref{H} (ii) and \eqref{10+}, we have \begin{equation}\label{17+} \begin{aligned} &\frac{1}{2}\int_{\Omega\backslash\Omega_{\delta/2}}| \nabla v_{\sigma} |^{p-2}\nabla v_{\sigma} \nabla \tau\phi\,dx\\ &\geq \frac{1}{2}\int_{\Omega\backslash\Omega_{\delta/2}} M(x)\gamma_0^{(\theta_0-1)(p-1)}\gamma_0^{p-1}\frac{{\zeta}_{\lambda, \gamma_0}(v_{\sigma}+\epsilon)}{(v_{\sigma}+\epsilon)^{p-1}}\tau\phi \\ &\geq \frac{1}{2}\int_{\Omega\backslash\Omega_{\delta/2}} M(x)\gamma_0^{(p-1)\theta_0}\frac{{\zeta}_{\lambda, \gamma_0}(v_{\sigma}+\epsilon)}{\gamma_0^{\theta_0(p-1)}}\tau\phi \\ &\geq \frac{1}{2}\int_{\Omega\backslash\Omega_{\delta/2}} M(x)[{f_ {\gamma_0}}(v_{\sigma}+\epsilon)+\lambda{g_ {\gamma_0}}(v_{\sigma}+\epsilon)]\tau\phi \\ &\geq \int_{\Omega\backslash\Omega_{\delta/2}}[a(x)f(v_{\sigma}+\epsilon)+\lambda b(x)g(v_{\sigma}+\epsilon)]\tau\phi, \end{aligned} \end{equation} for each $\lambda\in (0, \lambda^*)$, $\sigma \in (0, \tilde{\sigma})$, $\epsilon>0$. Now, denoting by \begin{equation}\label{v} v(x):=\lim_{\sigma\to 0}v_{\sigma}(x) =[\psi_\lambda(\omega_M(x))]^{\theta_0}, \quad x \in \overline{\Omega}, \end{equation} it follows from Lemma \ref{H} (iii), $v_{\sigma}>v>0$ in $\Omega\backslash\Omega_{\delta/2}$ and $ q \in (p-1, p]$ that \begin{equation}\label{18+} \begin{aligned} \Big[\frac{H_{\lambda,\gamma_0}(v_{\sigma})}{v_{\sigma}}\Big]^{q-(p-1)} &\leq \Big[\frac{H_{\lambda, \gamma_0}(v)}{v}\Big]^{q-(p-1)}\\ &\leq\big\|\frac{H_{\lambda, \gamma_0}(v)}{v} \big\|_{L^\infty(\Omega\backslash\Omega_{\delta/2})}^{q-(p-1)}\\ &= C_2\Big[\min_{\overline{\Omega}\backslash\Omega_{\delta/2}} v\Big]^{\frac{(\theta_0-1)(p-1-q)}{\theta_0}}\\ &\leq C_2 v^{\frac{(\theta_0-1)(p-1-q)}{\theta_0}}\\ &< C_2 v_{\sigma}^{\frac{(\theta_0-1)(p-1-q)}{\theta_0}}, \quad \text{for all } x \in \overline{\Omega\backslash\Omega}_{\delta/2}, \end{aligned} \end{equation} where \[ C_2={\big\|\frac{H_{\lambda,\gamma_0}(v)}{v}\big\|_{L^\infty(\Omega \backslash\Omega_{\delta/2})}^{q-(p-1)}} \big/{\Big[\min_{\overline{\Omega}\backslash\Omega_{\delta/2}} v\Big]^{\frac{(\theta_0-1)(p-1-q)}{\theta_0}}}>0 \] is independent of $\sigma$. Now we show that \begin{align*} \frac{1}{2}\int_{\Omega\backslash\Omega_{\delta/2}}| \nabla v_{\sigma} |^{p-2}\nabla v_{\sigma} \nabla \tau\phi\,dx &\geq \frac{\gamma_0^{(p-1)(\theta_0-1)}[{f}(\gamma_0)+\lambda {g}(\gamma_0)]}{4}\int_{\Omega\backslash\Omega_{\delta/2}} \beta(x)\tau\phi\,dx\\ &\quad +\frac{\gamma_0^{p-1-q}}{4 C_2\|\nabla\omega_M\|^q_{L^\infty(\Omega)}} \int_{\Omega\backslash\Omega_{\delta/2}}\alpha(x)|\nabla v_{\sigma} |^q]\tau\phi\,dx \end{align*} and as a consequence of this, using \eqref{mu'}, we obtain $$ \frac{1}{2}\int_{\Omega\backslash\Omega_{\delta/2}}| \nabla v_{\sigma} |^{p-2}\nabla v_{\sigma} \nabla \tau\phi\,dx \geq\mu\int_{\Omega\backslash\Omega_{\delta/2}} V(x, \nabla v_{\sigma})\tau\phi\,dx $$ for each $0\leq\mu<\mu^*_\lambda$. By \eqref{18+} and Lemma \ref{H} (iii)-(iv) in \eqref{16++}, we have \begin{align*}%\label{comp2} &\frac{1}{2}\int_{\Omega\backslash\Omega_{\delta/2}}| \nabla v_{\sigma} |^{p-2}\nabla v_{\sigma} \nabla \tau\phi\,dx\\ &\geq\frac{1}{2}\int_{\Omega\backslash\Omega_{\delta/2}} M(x) v_{\sigma}^{\frac{(\theta_0-1)(p-1)}{\theta_0}}\gamma_0^{p-1} \Big[\frac{H_{\lambda, \gamma_0}(v_{\sigma})}{v_{\sigma}}\Big]^{p-1}\tau\phi \\ &\geq \frac{1}{4}\int_{\Omega\backslash\Omega_{\delta/2}} M(x)\gamma_0^{(\theta_0-1)(p-1)}\gamma_0^{p-1} \Big[\frac{f(\gamma_0)}{\gamma_0^{p-1}}+\lambda \frac{g(\gamma_0)}{\gamma_0^{p-1}}\Big]\tau\phi \\ &\quad +\frac{1}{4}\int_{\Omega\backslash\Omega_{\delta/2}} M(x)v_{\sigma}^{\frac{(\theta_0-1)(p-1-q)}{\theta_0}}\gamma_0^{p-1-q} \Big[\frac{H_{\lambda, \gamma_0}(v_{\sigma})}{v_{\sigma}}\Big]^{p-1} \frac{\theta_0^q}{\theta_0^q} v_{\sigma}^{\frac{(\theta_0-1)q}{\theta_0}} \gamma_0^q \tau\phi \\ &\geq \frac{\gamma_0^{(p-1)\theta_0}[\frac{f(\gamma_0)}{\gamma_0^{p-1}}+\lambda \frac{g(\gamma_0)}{\gamma_0^{p-1}}]}{4}\int_{\Omega\backslash\Omega_{\delta/2}} M(x)\tau\phi \\ &\quad +\frac{\gamma_0^{p-1-q}}{4 C_2}\int_{\Omega\backslash\Omega_{\delta/2}} M(x)\theta_0^q v_{\sigma}^{\frac{(\theta_0-1)q}{\theta_0}}\frac{\gamma_0^q}{\theta_0^q} \Big[\frac{H_{\lambda, \gamma_0}(v_{\sigma})}{v_{\sigma}}\Big]^q\tau\phi. \end{align*} Using \eqref{mu'}, Lemma \ref{eta} (iv), the definition of $M$ and (V1), we obtain \begin{equation}\label{19++} \begin{aligned} &\frac{1}{2}\int_{\Omega\backslash\Omega_{\delta/2}} | \nabla v_{\sigma} |^{p-2}\nabla v_{\sigma} \nabla \tau\phi\,dx\\ &\geq\frac{\gamma_0^{(p-1)(\theta_0-1)}[{f}(\gamma_0)+\lambda {g}(\gamma_0)]}{4}\int_{\Omega\backslash\Omega_{\delta/2}} M(x)\tau\phi\,dx\\ &\quad +\mu^*_\lambda(\gamma_0)\|\nabla\omega_M\|_{L^\infty(\Omega)}^q \int_{\Omega\backslash\Omega_{\delta/2}} M(x)[\theta_0\psi_\lambda(\omega_M+\sigma)^{\theta_0-1} \psi'_\lambda(\omega_M+\sigma)]^q\tau\phi\,dx\\ &\geq \frac{\gamma_0^{(p-1)(\theta_0-1)}[{f}(\gamma_0)+\lambda {g}(\gamma_0)]}{4}\int_{\Omega\backslash\Omega_{\delta/2}} \beta(x)\tau\phi\,dx\\ &\quad +\mu^*_\lambda(\gamma_0) \int_{\Omega\backslash\Omega_{\delta/2}}\alpha(x) [\theta_0\psi_\lambda(\omega_M+\sigma)^{\theta_0-1}\psi'_\lambda(\omega_M+\sigma) |\nabla\omega_M|]^q \tau\phi\,dx\\ &\geq \mu^*_\lambda(\gamma_0)\int_{\Omega\backslash \Omega_{\delta/2}}[\beta(x)+\alpha(x)|\nabla v_{\sigma} |^q]\tau\phi\,dx\\ &\geq \mu\int_{\Omega\backslash\Omega_{\delta/2}} V(x, \nabla v_{\sigma})\tau\phi\,dx. \end{aligned} \end{equation} Going back to \eqref{16++} and using \eqref{17+} and \eqref{19++}, we obtain \begin{equation}\label{21+} \begin{aligned} &\int_{\Omega\backslash\Omega_{\delta/2}}|\nabla v_{\sigma} |^{p-2}\nabla v_{\sigma} \nabla \tau\phi\,dx\\ &\geq\int_{\Omega\backslash\Omega_{\delta/2}}[a(x)f( v_{\sigma}+\epsilon)+\lambda b(x)g( v_{\sigma}+\epsilon)+\mu V(x, \nabla v_{\sigma})]\tau\phi, \end{aligned} \end{equation} for each $0<\lambda<\lambda^*$, $0\leq\mu<\mu^*_\lambda$, $\epsilon>0$. Below we work on the ring $\Omega_\delta$. As before, using the definition of $v_{\sigma}$, it follows that \begin{equation}\label{21++} \begin{aligned} &\int_{\Omega_\delta} |\nabla v_{\sigma} |^{p-2}\nabla v_{\sigma} \nabla (1-\tau)\phi\,dx\\ &= \int_{\Omega_\delta}|\nabla\omega_M|^{p-2} \nabla\omega_M\nabla\big\{{\theta_0}^{p-1} [\psi_\lambda(\omega_M+\sigma)]^{(\theta_0-1)(p-1)}\\ &\quad\times [\psi'_\lambda(\omega_M+\sigma)]^{p-1}(1-\tau)\phi\big\} \\ &\quad -(\theta_0-1)(p-1)\int_{\Omega_\delta}|\nabla\omega_M|^{p} {\theta_0}^{p-1}[\psi_\lambda(\omega_M+\sigma)]^{(\theta_0-1)(p-1)-1}\\ &\quad\times [\psi'_\lambda(\omega_M+\sigma)]^{p}(1-\tau)\phi \\ &\quad -(p-1)\int_{\Omega_\delta}|\nabla\omega_M|^{p} {\theta_0}^{p-1}[\psi_\lambda(\omega_M+\sigma)]^{(\theta_0-1)(p-1)}\\ &\quad [\psi'_\lambda(\omega_M+\sigma)]^{p-2} \psi''_\lambda(\omega_M+\sigma)(1-\tau)\phi. \end{aligned} \end{equation} In a way similar to the one for \eqref{17+}, we have \begin{equation}\label{22+} \begin{aligned} &\frac{1}{2}\int_{\Omega_\delta}|\nabla v_{\sigma} |^{p-2}\nabla v_{\sigma} \nabla (1-\tau)\phi\,dx\\ &\geq \frac{{\theta_0}^{p-1}}{2}\int_{\Omega_\delta} M(x) [\psi_\lambda(\omega_M+\sigma)]^{(\theta_0-1)(p-1)} [\psi'_\lambda(\omega_M+\sigma)]^{p-1}(1-\tau)\phi\,dx\\ &\geq \int_{\Omega_\delta}[a(x)f(v_{\sigma}+\epsilon) +\lambda b(x)g(v_{\sigma}+\epsilon)](1-\tau)\phi\,dx, \end{aligned} \end{equation} for each $\lambda \in (0, \lambda^*)$, $\sigma \in (0,\tilde{\sigma})$, $\epsilon>0$. Besides this, we will show that \begin{align*} &\frac{1}{2}\int_{\Omega_\delta}| \nabla v_{\sigma} |^{p-2}\nabla v_{\sigma} \nabla (1-\tau)\phi\,dx\\ &\geq\frac{(1-\theta_0)(p-1)[\gamma_0 H_{\lambda, \gamma_0}(1)]^{p-q}}{4\Vert \alpha \Vert_{\infty}}\int_{\Omega_\delta}\alpha(x)|\nabla v_\sigma|^q(1-\tau)\phi\,dx\\ &\quad +\frac{\gamma_0^{(p-1)(\theta_0-1)}[{f}(\gamma_0)+\lambda {g}(\gamma_0)]}{4}\int_{\Omega_\delta}\beta(x)(1-\tau)\phi\,dx \end{align*} and as a consequence of this, using \eqref{mu'}, we obtain $$ \frac{1}{2}\int_{\Omega_\delta}| \nabla v_{\sigma} |^{p-2}\nabla v_{\sigma} \nabla (1-\tau)\phi\,dx\geq\mu\int_{\Omega_\delta}V(x, \nabla v_{\sigma})(1-\tau)\phi\,dx $$ for each $0\leq \mu<\mu^*_\lambda$. In fact, from the properties of the auxiliary functions and (M1), we have \begin{align*} &\frac{1}{2}\int_{\Omega_\delta}| \nabla v_{\sigma} |^{p-2}\nabla v_{\sigma} \nabla (1-\tau)\phi\,dx\\ &\geq -\frac{(\theta_0-1)(p-1)}{4}\int_{\Omega_\delta} |\nabla\omega_M|^p\theta_0^{p-1}[\psi_\lambda(\omega_M+\sigma) ]^{(\theta_0-1)(p-1)-1}\\ &\quad\times [\psi'_\lambda(\omega_M+\sigma)]^p(1-\tau)\phi\\ &\quad +\frac{1}{4}\int_{\Omega_\delta}|\nabla\omega_M|^{p-2} \nabla\omega_M\nabla\{ {\theta_0}^{p-1}[\psi_\lambda(\omega_M+\sigma)]^{(\theta_0-1)(p-1)}\\ &\quad\times [\psi'_\lambda(\omega_M+\sigma)]^{p-1}(1-\tau)\phi\}\\ &= \frac{(1-\theta_0)(p-1)}{4}\int_{\Omega_\delta} |\nabla\omega_M|^p\theta_0^{p-1}v_{\sigma}^{\frac{p(\theta_0-1) -\theta_0}{\theta_0}}\frac{\gamma_0^p}{\theta_0^p} \Big[\frac{H_{\lambda, \gamma_0}(v_{\sigma})}{v_{\sigma}}\Big]^p(1-\tau)\phi\\ &\quad +\frac{1}{4}\int_{\Omega_\delta}M(x){\theta_0}^{p-1} [\psi_\lambda(\omega_M+\sigma)]^{(\theta_0-1)(p-1)} [\psi'_\lambda(\omega_M+\sigma)]^{p-1}(1-\tau)\phi. \end{align*} That is, from \eqref{1.27} and Lemma \ref{H}, we have \begin{equation}\label{aa} \begin{aligned} &\frac{1}{2}\int_{\Omega_\delta}| \nabla v_{\sigma} |^{p-2}\nabla v_{\sigma} \nabla (1-\tau)\phi\,dx\\ &=\frac{(1-\theta_0)(p-1)}{4}\int_{\Omega_\delta} \frac{|\nabla\omega_M|^p}{v_{\sigma}}\frac{\theta_0^{p-1}}{\theta_0^{p-1}} v_{\sigma}^{\frac{p(\theta_0-1)}{\theta_0}} \gamma_0^p\Big[\frac{H_{\lambda, \gamma_0}(v_{\sigma})}{v_{\sigma}}\Big]^p(1-\tau)\phi\,dx\\ &\quad +\frac{1}{4}\int_{\Omega_\delta}M(x){\theta_0}^{p-1} v_{\sigma}^{\frac{(\theta_0-1)(p-1)}{\theta_0}}\frac{\gamma_0^{p-1}}{\theta_0^{p-1}} \Big[\frac{H_{\lambda, \gamma_0}(v_{\sigma})}{v_{\sigma}}\Big]^p(1-\tau)\phi\,dx\\ &\geq \frac{(1-\theta_0)(p-1)}{4}\int_{\Omega_\delta}|\nabla\omega_M|^q v_{\sigma}^{\frac{p(\theta_0-1)}{\theta_0}} \gamma_0^p\Big[\frac{H_{\lambda, \gamma_0}(v_{\sigma})}{v_{\sigma}}\Big]^p(1-\tau)\phi\,dx\\ &\quad + \frac{1}{4}\int_{\Omega_\delta}M(x)\gamma_0^{(\theta_0-1)(p-1)} \gamma_0^{p-1}\Big[\frac{f(\gamma_0)}{\gamma_0^{p-1}}+\lambda \frac{g(\gamma_0)}{\gamma_0^{p-1}}\Big](1-\tau)\phi\,dx. \end{aligned} \end{equation} Using that $v_{\sigma}<1$ in $\Omega_\delta$ (see \eqref{d}), $q[v_{\sigma}(x) ]^{\frac{(\theta_0-1)q}{\theta_0}}, \quad \text{for each } x \in \Omega_\delta \end{equation} and from Lemma \ref{H}, we have \begin{equation}\label{24+} [H_{\lambda, \gamma_0}(1)]^{p-q}\Big[\frac{H_{\lambda, \gamma_0}(v_{\sigma})}{v_{\sigma}}\Big]^q\leq \Big[\frac{H_{\lambda, \gamma_0}(v_{\sigma})}{v_{\sigma}}\Big]^p, \ x \in \Omega_\delta. \end{equation} From \eqref{23+} and \eqref{24+}, we rewrite \eqref{aa} as \begin{align*} &\frac{1}{2}\int_{\Omega_\delta}|\nabla v_{\sigma} |^{p-2}\nabla v_{\sigma} \nabla (1-\tau)\phi\,dx\\ &\geq \frac{(1-\theta_0)(p-1)}{4}\int_{\Omega_\delta}|\nabla\omega_M|^q v_{\sigma}^{\frac{(\theta_0-1)q}{\theta_0}}\gamma_0^{p-q}\gamma_0^q \frac{\theta_0^q}{\theta_0^q}[H_{\lambda, \gamma_0}(1)]^{p-q}\\ &\quad\times \Big[\frac{H_{\lambda,\gamma_0}(v_{\sigma})}{v_{\sigma}} \Big]^q(1-\tau)\phi\,dx +\frac{\gamma_0^{(p-1)\theta_0}[\frac{f(\gamma_0)}{\gamma_0^{p-1}}+\lambda \frac{g(\gamma_0)}{\gamma_0^{p-1}}]}{4}\int_{\Omega_\delta} M(x)(1-\tau)\phi\,dx\\ &= \frac{(1-\theta_0)(p-1)[\gamma_0 H_{\lambda, \gamma_0}(1)]^{p-q}}{4\Vert \alpha \Vert_{\infty}}\int_{\Omega_\delta}\Vert \alpha \Vert_{\infty}\theta_0^q v_{\sigma}^{\frac{(\theta_0-1)q}{\theta_0}} \Big[\frac{\gamma_0 H_{\lambda, \gamma_0}(v_{\sigma})}{\theta_0 v_{\sigma}}\Big]^q \\ &\quad\times |\nabla\omega_M|^q(1-\tau)\phi +\frac{\gamma_0^{(p-1)(\theta_0-1)}[{f}(\gamma_0)+\lambda {g}(\gamma_0)]}{4}\int_{\Omega_\delta} M(x)(1-\tau)\phi\,dx. \end{align*} From Lemma \ref{eta} (iv), \eqref{mu'} and definitions of $v_{\sigma}$, $M$ and (V1), we obtain \begin{equation}\label{25+} \begin{aligned} &\frac{1}{2}\int_{\Omega_\delta}|\nabla v_{\sigma}|^{p-2}\nabla v_{\sigma} \nabla (1-\tau)\phi\,dx\\ &\geq \mu^*_\lambda(\gamma_0)\|\alpha\|_{L^\infty(\Omega)}\int_{\Omega_\delta}| \nabla v_{\sigma}|^q(1-\tau)\phi\,dx +\mu^*_\lambda(\gamma_0)\int_{\Omega_\delta} M(x)(1-\tau)\phi \\ &\geq \mu^*_\lambda(\gamma_0)\int_{\Omega_\delta}[\alpha(x)|\nabla v_{\sigma} |^q+\beta(x)](1-\tau)\phi\,dx\\ &\geq \mu\int_{\Omega_\delta}V(x, \nabla v_{\sigma})(1-\tau)\phi\,dx. \end{aligned} \end{equation} for each $0\leq\mu<\mu^*_\lambda$. Considering \eqref{21++} and using \eqref{22+} and \eqref{25+}, we have \begin{equation}\label{26+} \begin{aligned} &\int_{\Omega_\delta}|\nabla v_{\sigma} |^{p-2}\nabla v_{\sigma} \nabla (1-\tau)\phi \\ &\geq\int_{\Omega_\delta}[a(x)f( v_{\sigma}+\epsilon)+\lambda b(x)g(v_{\sigma}+\epsilon)+\mu V(x, \nabla v_{\sigma})](1-\tau)\phi , \end{aligned} \end{equation} for each $0<\lambda<\lambda^*$, $0\leq \mu<\mu^*_\lambda$, $0 <\sigma <\tilde{\sigma}$ and $\epsilon>0$. Therefore, replacing \eqref{21+} and \eqref{26+} in \eqref{15+}, we conclude the proof of the existence of a upper solution for Proposition \ref{SuperDL}. To finalize the proof of the proposition, we need to verify the estimate for $\mu^*$. Assume $(F_0)$. So, from Lemma \ref{Gamma2} (ii), we have $$ \lim_{\gamma\to 0}\Gamma_{0, 1}(\gamma) ={f_0^{\frac{-1}{p-1}}}>\|\omega_M\|_{L^\infty(\Omega)} $$ and a consequence of this there exists a $\tilde{\gamma}>0$ sufficiently small such that $(0, \tilde{\gamma})\subset \mathcal{A}$. Given $0<\lambda<\lambda^0$, where $\lambda^0=\lambda^i$ with $i=0$ ($\lambda^i$ was defined in Theorem \ref{DL+}), we claim that there exists a $\gamma_0<\tilde{\gamma}$ such that $\lambda<\Lambda^*(\gamma)$ for all $0<\gamma<\gamma_0$. In fact, from $\lambda<\lambda^0$ and Lemma \ref{Gamma2} (ii) we have $$ \lim_{\gamma\to 0}\Gamma_{\lambda,1}(\gamma)=\frac{1}{(f_0+\lambda g_0)^{\frac{1}{p-1}}}>\|\omega_M\|_{L^\infty(\Omega)}. $$ So, there exists a $\gamma_0<\tilde{\gamma}$ such that $\Gamma_{\lambda,1}(\gamma)>\|\omega_M\|_{L^\infty(\Omega)} =\Gamma_{\Lambda^*(\gamma), 1}(\gamma)$ for $0<\gamma<\gamma_0$. Now, by Lemma \ref{Gamma2} (iii), we obtain $\lambda<\Lambda^*(\gamma)$, for all $\gamma \in (0, \gamma_0)$. From \eqref{ex} and \eqref{ex1} we have \begin{align*} \mu^*_{\lambda} &\geq \sup\{\mu^*_{\lambda}(\gamma):\gamma \in (0, \gamma_0) ~ \text{and} ~ \lambda<\Lambda^*(\gamma)\}\\ &\geq \liminf_{\gamma\to 0} \mu^*_{\lambda}(\gamma)=\min\Big\{\frac{[f^0+\lambda g^0]^{p-1-q}}{4\|\nabla\omega_M\|_{L^\infty(\Omega)}^q}, \frac{f^0+\lambda g^0}{4}\Big\}. \end{align*} If $(F_{\infty})$ occurs, we proceed in a similar manner to the above case. We point out that, in this case, $\gamma_0$ is large. This completes the proof of Proposition \ref{SuperDL}. \end{proof} \section{Conclusion of the proof of Theorem \ref{DL+}} We begin by constructing a lower solution for problem \eqref{8}. It follows from the definition of $\lambda_*$ that given $\lambda>\lambda_*$, there exists a $0<\epsilon_1 \leq \min\{\gamma_0,\gamma_0^{\theta_0}\}$ such that $$ f(s) + \lambda g(s) \geq \lambda_{ \Omega}(\rho) s^{p-1},\quad \text{for } 00$ such that $C\|\varphi_\Omega\|_{L^\infty(\Omega)}={\epsilon_1}/2$, it follows that \begin{equation}\label{27ii} C\|\varphi_\Omega\|_{L^\infty(\Omega)}+\epsilon 0$ is the eigenfunction associated to the first eigenvalue $\lambda_{\Omega}>0$ of problem \eqref{7}. Thus, given $\phi \in C_0^\infty(\Omega)$ with $\phi\geq 0$, we obtain $$ \int_\Omega|\nabla (C\varphi_\Omega)|^{p-2} \nabla (C\varphi_\Omega) \nabla\phi\,dx\leq\int_\Omega[\lambda b(x)g( C\varphi_\Omega+\epsilon) +a(x)f( C\varphi_\Omega+\epsilon)]\phi\,dx; $$ that is, $C\varphi_\Omega$ is a lower solution of \eqref{8} for each $0<\epsilon<\epsilon_1/{2}$, $0<\lambda<\lambda^*$ and $0<\mu < \mu^*_\lambda$, because of the positivity of $V$. Now, we claim that \begin{equation}\label{compp} C\varphi(x)\leq v_{\sigma}(x), \quad x \in \overline{\Omega}. \end{equation} First, we consider $q \in [0, p-1]$. In this case, $v_{\sigma}=\psi_\lambda(\omega_M+\sigma)$ is defined in \eqref{17}. So, from \eqref{27ii} and \eqref{12}, for all $\phi \in C_0^\infty(\Omega)$, $\phi\geq 0$, we have \begin{equation}\label{22'} \begin{aligned} &\int_\Omega|\nabla (C\varphi_\Omega)|^{p-2} \nabla (C\varphi_\Omega) \nabla\phi\,dx\\ &\leq \int_\Omega\Big[\gamma_0^{p-1} a(x)\frac{f(C\varphi_\Omega+\epsilon)}{(C\varphi_\Omega+\epsilon)^{p-1}} +\gamma_0^{p-1}\lambda b(x)\frac{g(C\varphi_\Omega+\epsilon)}{(C\varphi_\Omega+\epsilon)^{p-1}}\Big]\phi\,dx \\ &\leq \int_\Omega\Big[\gamma_0^{p-1} a(x)\frac{{f_{ \gamma_0}}(C\varphi_\Omega+\epsilon)}{(C\varphi_\Omega +\epsilon)^{p-1}}+\gamma_0^{p-1}\lambda b(x)\frac{{g_ {\gamma_0}}(C\varphi_\Omega+\epsilon)} {(C\varphi_\Omega+\epsilon)^{p-1}}\Big]\phi\,dx\\ &=\int_\Omega M(x)\gamma_0^{p-1}\frac{{\zeta}_{\lambda, \gamma_0}(C\varphi_\Omega+\epsilon)}{(C\varphi_\Omega+\epsilon)^{p-1}}\phi\,dx\\ &\leq \int_\Omega M(x)\gamma_0^{p-1}\frac{{\zeta}_{\lambda, \gamma_0}(C\varphi_\Omega)}{(C\varphi_\Omega)^{p-1}}\phi\,dx. \end{aligned} \end{equation} Moreover, from \eqref{19} and Lemma \ref{H}, we have \begin{equation}\label{23'} \int_\Omega|\nabla v_\sigma)|^{p-2} \nabla v_\sigma \nabla\phi\,dx \geq \int_\Omega M(x)\gamma_0^{p-1}\frac{{\zeta}_{\lambda, \gamma_0} (v_{\sigma})}{v_{\sigma}^{p-1}}\phi\,dx, \end{equation} for all $\phi \in C_0^\infty(\Omega), \phi\geq 0$. So, from \eqref{22'}, \eqref{23'}, ${\zeta}_{\lambda, \gamma_0}(s)/{s^{p-1}}$ non-increasing in $s>0$ and $C\varphi_\Omega=0<\psi_{\lambda}(\sigma)=v_{\sigma}$ on $\partial\Omega$, we apply a comparison principle for weak solutions (see Tolksdorf \cite{Tolks1}) to obtain \eqref{compp}. In the second case, that is $q \in (p-1,p]$, we recall that $v_{\sigma}=[\psi_\lambda(\omega_M+\sigma)]^{\theta_0}$, where $\theta_0 \in (\Vert w_M \Vert_{\infty}f_i^{1/(p-1)},1)$, see \eqref{defv}. In a similar way to the first case (that is, $q \in [0,p-1]$), we obtain \begin{equation}\label{27+i} \int_\Omega|\nabla v_\sigma|^{p-2} \nabla v_\sigma \nabla\phi\,dx\geq \int_\Omega M(x)\gamma_0^{(p-1)\theta_0}\frac{{\zeta}_{\lambda, \gamma_0}(v_{\sigma})}{v_{\sigma}^{p-1}}\phi\,dx, \end{equation} for all $\phi \in C_0^\infty(\Omega)$ with $\phi\geq 0$. From \eqref{27ii}, definitions and properties of auxiliary functions $\xi_{f,\gamma_0}$, $\xi_{g,\gamma_0}$ and $\xi_{\lambda,\gamma_0}$, we have \begin{equation} \label{27+ii} \begin{aligned} &\int_\Omega|\nabla (C\varphi_\Omega)|^{p-2} \nabla (C\varphi_\Omega) \nabla\phi\,dx\\ &\leq \int_\Omega\left[\gamma_0^{(p-1)\theta_0} a(x)\frac{f(C\varphi_\Omega+\epsilon)}{(C\varphi_\Omega+\epsilon)^{p-1}} +\gamma_0^{(p-1)\theta_0}\lambda b(x)\frac{g(C\varphi_\Omega+\epsilon)} {(C\varphi_\Omega+\epsilon)^{p-1}}\right]\phi\,dx\\ &\leq \int_\Omega\Big[\gamma_0^{(p-1)\theta_0}a(x)\frac{f_{ \gamma_0}(C\varphi_\Omega+\epsilon)}{(C\varphi_\Omega +\epsilon)^{p-1}}+\gamma_0^{(p-1)\theta_0}\lambda b(x)\frac{{g_ {\gamma_0}}(C\varphi_\Omega+\epsilon)}{(C\varphi_\Omega+\epsilon)^{p-1}} \Big]\phi\,dx\\ &=\int_\Omega M(x)\gamma_0^{(p-1)\theta_0}\frac{{\zeta}_{\lambda, \gamma_0}(C\varphi_\Omega+\epsilon)}{(C\varphi_\Omega+\epsilon)^{p-1}}\phi\,dx\\ &\leq\int_\Omega M(x)\gamma_0^{(p-1)\theta_0}\frac{{\zeta}_{\lambda, \gamma_0}(C\varphi_\Omega)}{(C\varphi_\Omega)^{p-1}}\phi\,dx, \end{aligned} \end{equation} for all $\phi \in C_0^\infty(\Omega), \phi\geq 0$. Hence, from \eqref{27+i}, \eqref{27+ii}, ${\zeta}_{\lambda, \gamma_0}(s)/{s^{p-1}}$ non-increasing in $s>0$ and $C\varphi_\Omega=0<\psi_{\lambda}^{\theta_0}(\sigma)=v_{\sigma}$ on $\partial\Omega$, the claim follows. Here, again we used Tolksdorf \cite{Tolks1}. Now, by taking $\sigma=1/m$ and $\epsilon=1/n$ with sufficiently large $m,n \in \mathbb{N}$, it follows from the lower upper solution Theorem (see Boccardo, Murat and Puel \cite{BMP}) that there exists $u_{m,n} \in W_0^{1, p}(\Omega)\cap L^\infty(\Omega)$ with $0 0 \quad \text{in } \mathbb{R}^N \quad \text{and $u\to 0$ as $|x|\to \infty$}, \end{gathered} \end{equation} for $10$ such that for each $0<\lambda<\lambda^*$ and $\epsilon>0$, there exist a $\mu^*=\mu^*_\lambda> 0$ and a function $ v=v_{\lambda, \mu} \in C^1(\mathbb{R}^N)$, both independent of $\epsilon$, with $v$ being a solution of $\eqref{SP+}$ for each $0\leq\mu<\mu^*$. Additionally: \begin{itemize} \item [(i)] if $(F_i)$ occurs for $i \in \{0,\infty\}$, then $$ \lambda^*\geq \frac{1}{g_i}\Big(\frac{1}{\|\omega_M\|_{L^\infty(\mathbb{R}^N)}^{p-1}} -f_i\Big); $$ \item[(ii)] there is $d>0$ such that $$ \mu^*_\lambda \geq d \min\big\{ {[f^i+\lambda g^i]^{\frac{p-1-q}{p-1}}}, {f^i+\lambda g^i}\big\}. $$ \end{itemize} \end{proposition} \begin{proof} The proof of this result is analogous to the proof of part one of Proposition \ref{SuperDL}. Considering $\Omega=\mathbb{R}^N$ and $\theta_0=1$, we define the set $\mathcal{A}=\mathcal{A}_{\mathbb{R}^N}=\{\gamma \in (0, \infty): \Gamma_{0, 1}(\gamma)>\|\omega_M\|_{L^\infty(\mathbb{R}^N)}\}$. So, we obtain \eqref{14} and the positive number $\lambda^*=\lambda^*(\mathbb{R}^N)=\sup\{\Lambda^*(\gamma): \gamma \in \mathcal{A}_{\mathbb{R}^N}\} $. Moreover, we define the positive number \begin{equation}\label{ex2} \mu^*_{\lambda}({\gamma})=\mu^*_{\lambda, \mathbb{R}^N}({\gamma})=\min\Big\{\frac{[{f}({\gamma})+\lambda {g}({\gamma})]^{\frac{p-1-q}{p-1}}} {4\|\nabla\omega_M\|_{L^\infty(\mathbb{R}^N)}^q}, \frac{{f}({\gamma})+\lambda {g}({\gamma})}{4}\Big\}. \end{equation} for each ${\gamma}, \lambda>0$. Now, for $0<\lambda<\lambda^*$, we take the number $\mu^*_\lambda=\mu^*_{\lambda, \mathbb{R}^N}>0$ as defined in \eqref{ex1}. Thus, for $0\leq\mu<\mu^*_\lambda$ given, we have that there exists a $\gamma_0 \in \mathcal{A}$ such that $\lambda<\lambda^*(\gamma_0)$ and $\mu<\mu^*_\lambda(\gamma_0)$. Now, we fix this $\gamma_0$. So, given $0<\lambda<\lambda^*$, we define $$ v(x)=v_\lambda(x)=\psi_\lambda(\omega_M(x)), \quad x \in \mathbb{R}^N $$ and, as a consequence of the properties of $\psi_\lambda$, we obtain that $v \in C^1(\mathbb{R}^N)$, $v(x) \to 0$ as $\Vert x \Vert \to \infty$ and $00$, we have \begin{equation}\label{+0} \| v\|_{L^\infty(\mathbb{R}^N)}<\gamma_0-\epsilon. \end{equation} So, for each $\phi \in C_0^\infty(\mathbb{R}^N)$ with $\phi \geq 0$ given, we have (in a similar way to \eqref{19}) that \begin{equation}\label{+1} \int_{\mathbb{R}^N}|\nabla v|^{p-2}\nabla v \nabla\phi\,dx \geq\int_{\mathbb{R}^N} M(x)\gamma_0^{p-1}\Big[\frac{ H_{\lambda, \gamma_0}(\psi_\lambda(\omega_M))}{\psi_\lambda(\omega_M))}\Big]^{p-1}\phi\,dx. \end{equation} Below, we analyze the previous integral in two parts. First, we have \begin{align*} &\frac{1}{2}\int_{\mathbb{R}^N} M(x)\gamma_0^{p-1}\Big[\frac{ H_{\lambda, \gamma_0}(\psi_\lambda(\omega_M))}{\psi_\lambda(\omega_M)}\Big]^{p-1}\phi\,dx\\ &\geq\frac{1}{2}\int_{\mathbb{R}^N} M(x)\gamma_0^{p-1}\frac{{\zeta}_{\lambda, \gamma_0}(v)}{v^{p-1}} \phi\,dx\\ &\geq \frac{1}{2}\int_{\mathbb{R}^N} M(x)\gamma_0^{p-1}\frac{{\zeta}_{\lambda, \gamma_0}(v+\epsilon)}{(v+\epsilon)^{p-1}} \phi\,dx\\ &\geq \frac{1}{2}\int_{\mathbb{R}^N} M(x)[f_{ \gamma_0}(v+\epsilon)+\lambda g_ {\gamma_0}(v+\epsilon)]\phi\,dx. \end{align*} As a consequence of this, \eqref{+0}, definitions of $\zeta_{f,\gamma_0}$, $\zeta_{g,\gamma_0}$ and $M$, we have \begin{equation}\label{+2} \begin{aligned} &\frac{1}{2}\int_{\mathbb{R}^N} M(x)\gamma_0^{p-1}\Big[\frac{ H_{\lambda, \gamma_0}(\psi_\lambda(\omega_M))}{\psi_\lambda(\omega_M)}\Big]^{p-1}\phi\,dx\\ &\geq\int_{\mathbb{R}^N}[a(x)f(v+\epsilon)+\lambda b(x)g(v+\epsilon)]\phi\,dx. \end{aligned} \end{equation} For the other part, using the properties of the auxiliary functions and \eqref{ex2}, we have \begin{equation}\label{+5} \begin{aligned} &\frac{1}{2}\int_{\mathbb{R}^N} M(x)\gamma_0^{p-1}\Big[\frac{ H_{\lambda, \gamma_0}(\psi_\lambda(\omega_M))}{\psi_\lambda(\omega_M)}\Big]^{p-1}\phi\,dx\\ &\geq \frac{1}{4}\int_{\mathbb{R}^N} M(x)\gamma_0^{p-1}\Big[\frac{f(\gamma_0)}{\gamma_0^{p-1}}+\lambda \frac{g(\gamma_0)}{\gamma_0^{p-1}}\Big]\phi\,dx\\ &\quad +\frac{1}{4}\int_{\mathbb{R}^N} M(x)\gamma_0^{p-1-q}\Big[\frac{ H_{\lambda, \gamma_0}(\psi_\lambda(\omega_M))}{\psi_\lambda(\omega_M)}\Big]^{p-1-q} \Big[\frac{\gamma_0 H_{\lambda, \gamma_0}(\psi_\lambda(\omega_M))}{\psi_\lambda(\omega_M)}\Big]^q\phi\,dx\\ &\geq \frac{({f}(\gamma_0)+\lambda {g}(\gamma_0))}{4}\int_{\mathbb{R}^N} M(x)\phi\,dx \\ &\quad +\frac{1}{4}\int_{\mathbb{R}^N} M(x)\gamma_0^{p-1-q}\Big[\frac{f(\gamma_0)}{\gamma_0^{p-1}}+\lambda \frac{g(\gamma_0)}{\gamma_0^{p-1}}\Big]^{\frac{p-1-q}{p-1}}[\psi'_\lambda(\omega_M)]^q \phi\,dx\\ &\geq \mu^*_\lambda\int_{\mathbb{R}^N}\beta(x)\phi\,dx+\mu^*_\lambda\int_{\mathbb{R}^N} M(x)[\psi'_\lambda(\omega_M)]^q |\nabla\omega_M |^q \phi\,dx\\ &\geq \mu^*_\lambda\int_{\mathbb{R}^N}[\beta(x)+\alpha(x)| \nabla v |^q ]\phi\,dx \geq \mu\int_{\mathbb{R}^N} V(x, \nabla v)\phi\,dx, \end{aligned} \end{equation} for each $0\leq\mu<\mu^*_\lambda$. Hence, replacing \eqref{+2} and \eqref{+5} in \eqref{+1}, we get that $v$ satisfies \eqref{SP+}, for each $0<\lambda<\lambda^*$ and $0\leq\mu<\mu^*_\lambda$. The estimates given for $\lambda^*$ and $\mu^*_\lambda$ are obtained in a similar way as those of Proposition \ref{SuperDL}. This proves Proposition \ref{T02+}. \section{Conclusion of the proof of Theorem \ref{NL+}} First, we note that $\mathcal{A}_{\mathbb{R}^N}\subset \mathcal{A}_{B_R}$ for all $R\geq 1$. In fact, if $\gamma \in \mathcal{A}_{\mathbb{R}^N}$, then (using Lemma \ref{Gamma2} (iv)) we have $$ \Gamma_{0,\theta}(\gamma)> \|\omega_M\|_{L^\infty(\mathbb{R}^N)} \geq \|(\omega_M)_{\vert_{B_R}}\|_{L^\infty(B_R)},\quad \text{for all } R\geq 1; $$ that is, $ \gamma \in \mathcal{A}_{B_R}$. So, we obtain $$ \lambda^*(\mathbb{R}^N)=\sup\{\lambda^*(\gamma): \gamma \in \mathcal{A}_{\mathbb{R}^N}\}\leq \sup\{\lambda^*(\gamma): \gamma \in \mathcal{A}_{B_R}\}=\lambda^*(B_R) $$ for all $R\geq 1$. Concerning $\mu^*_\lambda$. As a direct consequence of \eqref{ex} and \eqref{ex2}, we obtain that $\mu^*_\lambda(\mathbb{R}^N) \leq \mu^*_\lambda(B_R)$, for all $R\geq 1$. So, given $\lambda_{*} < \lambda <\lambda^*(\mathbb{R}^N) $, $0 \leq \mu < \mu^{*}_\lambda(\mathbb{R}^N)$ and taking $v_R=v_{\vert_{B_R}}$ as an upper solution, there exists (Theorem \ref{DL+} and its demonstration) a $u_R\in W^{1,p}_0(B_R)\cap C(\overline{B}_R)$ with $0 0 \quad \text{in } B_R,\quad u_R = 0 \quad \text{on } \partial B_R, \end{gathered} \end{equation} for each $R> 1$, where $v$ is given by Proposition \ref{T02+}. Besides this, from the definition of $\lambda_{*}$, $0<\lambda< \lambda_{*}$ and $\lambda_1(\rho)=\lim_{R\to\infty}\lambda_{ B_R}(\rho)$, it follows that there exists a $L_0>1$ such that $\lambda_{ B_{L_0}}(\rho)<\lambda g_0+f_0.$ That is, from the monotonicity of the first eigenvalue concerning the domain, there exists one $\delta=\delta(L_0)>0$ such that \begin{equation}\label{02+} {f(s)+\lambda {g(s)}}> \lambda_{ B_{R}}(\rho) {s^{p-1}}, \quad \text{for all } s \in (0, \delta) \text{ and } R\geq L_0. \end{equation} Now, considering $C_{L_0}$ the constant of the lower solution of \eqref{01+} with $R={L_0}$ defined in \eqref{27ii}, we take a sufficiently small $C=C(\delta) \in (0, C_{L_0})$ such that \begin{equation}\label{05+} 0L_0. $$ Now, proceeding as in the end of proof of Theorem \ref{DL+}, we finish the proof of Theorem \ref{NL+}. \end{proof} \subsection*{Acknowledgments} M. C. Rezende was partially supported by CNPq/Brasil. C. A. 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